Answer
$ f^{-1}(x)=\frac{1}{2}cos^{-1}(\frac{x}{3})-\frac{1}{2} $
range of $f(x)$: $[-3,3]$
domain and range of $f^{-1}(x)$: $[-3,3]$ and $[-\frac{1}{2}-\frac{\pi}{4}, -\frac{1}{2}+\frac{\pi}{4}]$
Work Step by Step
Step 1. $f(x)=3sin(2x+1) \longrightarrow y=3sin(2x+1)\longrightarrow x=3sin(2y+1)\longrightarrow y=\frac{1}{2}cos^{-1}(\frac{x}{3})-\frac{1}{2} \longrightarrow f^{-1}(x)=\frac{1}{2}cos^{-1}(\frac{x}{3})-\frac{1}{2} $
Step 2. We can find the domain and range of $f(x)$: $[-\frac{1}{2}-\frac{\pi}{4}, -\frac{1}{2}+\frac{\pi}{4}]$ and $[-3,3]$
Step 3. We can find the domain and range of $f^{-1}(x)$: $[-3,3]$ and $[-\frac{1}{2}-\frac{\pi}{4}, -\frac{1}{2}+\frac{\pi}{4}]$
